Update yes, it was.
I changed
s/ //g;
to s/\W//g; thanks, Albannach! (wave)

#
#
#
use strict;
my %symbol;
my %tri;
my @trikeys;
my $line;
my $ctr;
print "Processing file...\n";
while ($line = <>) {
    for (split /\W/,$line) {
        #discard all non-alpha. be *greedy*        
        s/\W//g;
        (length($_)>2) && $symbol{lc($_)}++;
    }
}

print "Collecting trigraphs...\n";
foreach (keys %symbol){
    for ($ctr=0; $ctr <= (length($_)-3);$ctr++) {
        $tri{lc(substr ($_,$ctr,3))}+= $symbol{$_};  
    }
}

@trikeys = sort {$tri{$b} <=> $tri{$a}} keys %tri;
print "Total Trigraphs : ",$#trikeys,"\n";
print "Trigraph\tCount\n";
foreach (@trikeys) {
    print "$_\t$tri{$_}\n"; 
}
[download]

Cool. You have a tiny bug:

print "Total Trigraphs : ",$#trikeys,"\n";
[download]

This undercounts the trigraphs by 1.

Also, the s/\W//g; line is not doing anything. All the \W characters have already been discarded by the split.

I have a not-quite-brute-force decipherer, but it's not really what leitchn was looking for. It assumes that you still know where the word boundaries are, and then it uses a heuristically guided search based partly on letter frequency and partly on repeated letter patterns. For example, if it sees the ciphertext ABCDDEFGHIJA, it guesses that the word is either glassworking, stanniferous, or scaffoldings. (If it isn't one of these, it won't be able to solve the puzzle, because it doesn't know the words.)

Hee's the program that generates the pattern dictionary:

#!/usr/bin/perl
@DICTS = </usr/dict/*>;
# @DICTS = ('/usr/dict/words');
load_dictionary(@DICTS);
{ local $, = "\0";
  while (($pat, $words) = each %words) {
    print $pat, @$words, "\n";
  }
}

sub pattern {
  my ($w) = @_;
  my $n = 'A';
  while (my ($l) = $w =~ /([a-z])/) {
    $w =~ s/$l/$n/g;
    $n++;
  }
  $w;
}

sub load_dictionary {
  my $s = time;
  local @ARGV = @_;
  while (<>) {
    chomp;
    next unless /^[a-z]*$/;
    next if $is_word{$_}++;
    push @{$words{pattern($_)}}, $_;
  } continue {
    my $n = keys %is_word;
    print STDERR "$n words loaded.\n" if $n % 10000 == 0 && $n > 0;
  }
  my $e = time - $s;
  print STDERR "Elapsed time to load dictionary: $e.\n";
}
[download]

I'd really like to find a faster way to do the pattern() function.